\(0-1\) predictions in binary regression

• The predict method gives you a probability.
• But often you have to make a make a prediction of \(\hat{Y}\) itself.
• Will you follow a marketing lead? Do you believe candidate will get an \(A\) when he drinks 5 cups of coffee?
• You usually construct predictions of \(Y\) using a threshold: \[\hat{Y}=1 \iff \hat{p} > c,\] where \(\hat{p}\) is a predicted probability obtained using predict and \(\hat{Y}\) is the predicted value of \(Y\).
• For instance, you could decide to do the action if if \(\hat{p} > 0.5\).

Admission data (again)

import pandas as pd
import statsmodels.formula.api as smf
import matplotlib.pyplot as plt
admission = pd.read_csv("https://stats.idre.ucla.edu/stat/data/binary.csv")
admission.head()

Confusion matrix

The confusion matrix shows shows true positives, false negatives, false positives, and true negatives.

Confusion matrix for admission with c = 0.5

model.pred_table(threshold = 0.5)

array([[254.,  19.],
       [ 97.,  30.]])

• Question: How many true positives are there?
  • Answer: True positives happen when both \(\hat{Y}=1\) and \(Y=1\). Hence the correct answer is \(30\).
• Question: How many false negatives are there?
  • Answer: False negatives correspond to \(\hat{Y}=0\) and \(Y=1\). There are 97 of these.

Confusion matrix for admission

model.pred_table(threshold = 0.6)

array([[266.,   7.],
       [114.,  13.]])

model.pred_table(threshold = 0.5)

array([[254.,  19.],
       [ 97.,  30.]])

Confusion matrix for admission

• Question: How can you make sure the number of false positives is \(0\)?
  • Answer: Make the threshold \(c=1\)!

model.pred_table(threshold = 1)

array([[273.,   0.],
       [127.,   0.]])

True positive rate and false positive rate

\[ \text{true positive rate}=\frac{\text{true positives}}{\text{true positives} + \text{false negatives}} \] Also known as sensitivity, recall, and hit rate. Abbreviated as tpr.

\[ \text{false positive rate}=\frac{\text{false positives}}{\text{false positives} + \text{true negatives}} \]

Also known as fall-out. Abbreviated as fpr.

• Here \(TPR = \frac{30}{30 + 97} = 0.24\) and \(FPR = \frac{19}{19 + 254} = 0.07\).

Receiver operating characteristic curve

The receiver operating characteristic curve. Source: Wikipedia.

How to make these plots

from sklearn import metrics
fpr, tpr, _ = metrics.roc_curve(admission.admit, model.predict())
plt.plot(fpr, tpr)
plt.show()

The area under the curve (i)

The area under the curve (ii)

• The area under the curve (AUC) is the integral of the receiver operating characteristic curve.
  • \(0.5 \leq AUC \leq 1\).
  • \(AUC = 1\) if we can predict perfectly.
  • \(AUC = 0.5\) if we can’t predict at all.
• Extremely widespread!
• Since the AUC doesn’t penalize model complexity it shouldn’t be used to compare models!

The \(R^2\) for binary regression models

Remember the setup for the \(R^{2}\) for non-linear regression: \[ 1-\frac{\sum_{i=1}^{n}(y_{i}-f(x;a_{1},\ldots a_{p}))^{2}}{\sum_{i=1}^{n}(y_{i}-m)^{2}}\] In general we could write \[R^{2}=1-\frac{R(x)}{R(m)}\] Where

• \(R(m)\) measures the distance between our best predictions and the data \(y\), and
• \(R(x)\) measures the distance between our best predictions and the data \(y\) when we take \(x\) into account.

The log-likelihood and the \(R^2\)

We use the log-likelihood instead of the quadratic loss! So, letting \(p_i = f(x;a_{1},\ldots a_{p})\), define the fitted likelihood \[l(x) = R(x) = \sum_{i=1}^{n}-y_{i}\log \hat{p}_{i}-(1-y_{i})\log(1-\hat{p}_{i}).\]

\[l(m) = R(m) = \sum_{i=1}^{n}-y_{i}\log m-(1-y_{i})\log(1-m),\] is the log likelihood of the null model. It measures how well we can predict \(y\) when we don’t know any \(x\) at all.

McFadden’s \(R^2\)

Now define the pseudo-\(R^2\) or McFadden \(R^2\) as \[R^2_{\textrm{McFadden}} = 1 - \frac{l(x)}{l(m)}.\]

• Since McFadden’s \(R^2\) doesn’t penalize model complexity it shouldn’t be used to compare models!

An example

probit_fit = smf.probit("admit ~ gre + C(rank) + gpa", data = admission).fit()
logit_fit = smf.logit("admit ~ gre + C(rank) + gpa", data = admission).fit()

def rsq_mcfadden(fit):
    lower = fit.llnull
    upper = fit.llf
    return 1 - upper / lower

print(rsq_mcfadden(probit_fit), rsq_mcfadden(logit_fit))

Optimization terminated successfully.
         Current function value: 0.573016
         Iterations 5
Optimization terminated successfully.
         Current function value: 0.573147
         Iterations 6
0.08313059666890721 0.08292194470084713

• Thus the \(R^2\)s are \(0.0831\) and \(0.0829\). But we may also access McFadden’s \(R^2\) directly: print(probit_fit.prsquared, logit_fit.prsquared).

• Question: Can you compare these \(R^2\)s to the \(R^2\)s from linear regression?

  • Answer: No, as we are looking the likelihoods now, not the quadratic distance. Also, the “ordinary” \(R^2\) tends to be small for binary data.

McFadden’s \(R^2\) vs the ordinary \(R^2\)

The following picture, Figure 5 from Chapter 5 of McFadden’s Urban Travel Demand: A Behavioral Analysis (1996) illustrates the typical relationsship between the least squares \(R^2\) (on the \(y\)-axis) and the pseudo-\(R^2\) on the \(x\)-axis.

McFadden’s comparison if the ordinary R^2 and McFadden’s R^2. Source: McFadden, D. Urban Travel Demand: A Behavioral Analysis (1996)

Akaike’s information criterion (recap)

• Akaike’s information criterion (AIC), defined as \[\text{AIC} = 2p - 2l,\] where \(p\) is the number of estimated parameters and \(l\) is the log-likelihood at the estimated parameters.
• The smaller the AIC the better!
• We add \(2p\) since a larger number of parameters allows for more noise. It penalizes model complexity.
• Penalization of model complexity is always needed when comparing model with different number of parameters!
• There is a theoretical argument for the choice \(p\)! but that is beyond the scope of this course.

The adjusted \(R^2\)

• The \(R^2\) is good for evaluating how well be can predict the outcome given our covariates, but it’s not good for choosing between models.
• It doesn’t correct for the bias that occurs when using the same data both to estimate the model parameters and evaluating model fit.

import numpy as np
rng = np.random.default_rng(seed = 313)
p = 10
n = 100
x = rng.normal(0, 1, (n, p)) # think about y ~ 1 + x1 + x2 + ... + x_p
y = rng.normal(3, 2, n) # True R^2 is 0!

• The true \(R^2\) for y ~ x is \(0\) in this model. However, its expectation is not.
• Can show that \[R^2 \sim \textrm{Beta}\left(\frac{k-1}{2},\frac{n-k}{2}\right),\] where \(k\) is the number of parameters including the intercept. (That is, \(k = p + 1\).)
• Using the properties of the Beta distribution, we find that the expected value of \(R^2\) is \[E(R^2) = \frac{\frac{k-1}{2}}{\frac{k-1}{2} + \frac{n-k}{2}} = \frac{k-1}{n - 1}.\]
• This number might be quite large when the number of predictors is close to the number of observations!

Constructing the adjusted \(R^2\) (i)

• Recall the definition of the \(R^2\): \[R^2 = 1 - \frac{\textrm{Sum of squares with predictors}}{\textrm{Sum of squares without predictors}}\]

• We can show that \(\textrm{Sum of squares with predictors}\) is biased for its population value, the true sum of squares with predictors at our estimated regression coefficient. But we can correct for this bias!

• One can show that \[\frac{n}{n-p-1}E(\textrm{Sum of squares with predictors})\] equals the true population sum of squares with predictors. Here \(p\) is the number of estimated regression coefficients, minus the intercept.

Constructing the adjusted \(R^2\) (ii)

• Moreover, we can show that \[\frac{n}{n-1} E(\textrm{Sum of squares with predictors})\]equals the true, population value of the sum of squares without predictors.
• It follows that a reasonable corrected \(R^2\) is \[R_a^2 = 1 - \frac{\frac{n}{n-p-1} \textrm{Sum of squares with predictors}}{\frac{n}{n-1}\textrm{Sum of squares without predictors}}\]
• Rearranging this, we find that \[R_a ^ 2 = 1 - (1 - R^2) \frac{n-1}{n-p-1} .\]

Notes on the adjusted \(R^2\)

1. The adjusted \(R^2\), or \(R^2_a\), can be less than \(0\). Try to understand why.
2. We haven’t proved that the adjusted \(R^2\) squared is unbiased for the true \(R^2\). Do you think it is?
3. Can you devise a simulation study to explore this problem?

Comments

• These methods can be used together with automatic mechanisms for selecting covariates, e.g. backwards and forwards regression.
• The BIC (“Bayesian information criterion”) is similar to the AIC, but defined as \(k\log n - 2\log L\)! (The AIC is \(2k-2\log L\)).
• There are many different methods too, but adjusted \(R^2\), ANOVA and AIC are most popular.

Conditions for inference in regression

1. The linear relationsship is true, \(y_i=\beta^T x_i + \epsilon\).
2. No autocorrelaton. The errors \(\epsilon_i\) are independent of each other given \(x_i\). (This is important.)
3. Normally distributed errors. The errors are normally distributed. (Not very important.)
4. Homoskedastic errors. The variance of the errors should not depend on \(x_i\).

Residuals

• Defined as \(\hat{y} - y\). It’s common to look at plots of residuals.
• We often plot “residuals vs. fitted”, i.e. \(\hat{y} - y\) vs \(\hat{y}\)
• The plots should be without patterns!

Residuals: Heteroskedastic errors

rng = np.random.default_rng(seed = 313)
x = rng.uniform(size=100)
y = 1 + x + rng.normal(size=100) * (x + 0.5) ** 2
mod = smf.ols("y ~ x", data = {}).fit()
plt.scatter(mod.predict(), mod.resid)

<matplotlib.collections.PathCollection at 0x2125c9c2320>

• The residuals are smaller when the absolute value of the fitted observations is small!

• You can make such plots when dealing with multiple covariates too.

QQ plot of residuals

Source: Introduction to Linear Regression Analysis (Elizabeth Peck, Geoffrey Vining)

QQ plot: About \(n\)

• Small \(n<20\) often don’t yield useful quantile-quantile plots!

• Bu \(n>30\) usually suffices.

QQ plot: \(n = 100\), normal distribution

import numpy as np
import statsmodels.api as sm
import matplotlib.pyplot as plt
rng = np.random.default_rng(seed = 313)
x = rng.normal(size = 100)
sm.qqplot(x, line ='45')
plt.show()

QQ plot: \(n = 10\), normal distribution

import numpy as np
import statsmodels.api as sm
import matplotlib.pyplot as plt
rng = np.random.default_rng(seed = 3)
x = rng.normal(size = 10)
sm.qqplot(x, line ='45')
plt.show()

Looks like heavy tails!

QQ plot: Large \(n\), horrible distribution

rng = np.random.default_rng(seed = 3)
x = rng.standard_t(size = 50, df = 3)
sm.qqplot(x, line ='45')
plt.show()

You’re looking for seriously heavy tails!

QQ plot: Large \(n\), good distribution

rng = np.random.default_rng(seed = 3)
x = rng.uniform(size = 50, low = 0, high = 1)
sm.qqplot(x, line ='45')
plt.show()

Heteroskedasticity

import seaborn as sns
rng = np.random.default_rng(seed = 313)
x = rng.uniform(size=50)
y = 1 + x + rng.normal(size=50) * (x + 0.5) ** 2
data = pd.DataFrame({"x": x, "y": y})
sns.lmplot(data = data, x = "x", y = "y")

<seaborn.axisgrid.FacetGrid at 0x2125fa7da20>

Robust standard error’s using White (1980)

• Inferential procedures such as confidence intervals and p-values are invalid under heteroskedasticity.
• Us the argument cov_type='HC0' inside the fit methods for robust standard errors.

smf.ols("y ~ x", data = data).fit().pvalues

Intercept    0.018321
x            0.035364
dtype: float64

smf.ols("y ~ x", data = data).fit(cov_type='HC0').pvalues

Intercept    0.002156
x            0.028677
dtype: float64

White, H. (1980). A Heteroskedasticity-Consistent Covariance Matrix Estimator and a Direct Test for Heteroskedasticity. Econometrica, 48(4), 817–838. https://doi.org/10.2307/1912934

Telephone data

Number of international calls from Belgium, taken from the Belgian Statistical Survey, published by the Ministry of Economy.

import seaborn as sns
import rdatasets as rd
telef = rd.data("Robustbase", "telef")
sns.lmplot(data = telef, x = "Year", y = "Calls")

<seaborn.axisgrid.FacetGrid at 0x2125fa4b0a0>

• What’s happening here? Clerical errors! Such errors are abundant.

• Two ways to fix such problems:

  • (a) Remove the “bad” data. But this is not always feasible.

  • (b) Use robust regression methods. Such methods are – sometimes – able to automatically ignore data that should be ignored.

Using rlm

robust = smf.rlm(
  "Calls ~ Year", 
  data = telef, 
  M = sm.robust.norms.HuberT()).fit()

sns.lmplot(data = telef, x = "Year", y = "Calls")
plt.plot(telef["Year"], robust.predict(), color = "red")

Duncan prestige (here)

The Duncan data frame has 45 rows and 4 columns. Data on the prestige and other characteristics of 45 U. S. occupations in 1950.

prestige = sm.datasets.get_rdataset("Duncan", "carData", cache=True).data
prestige.head()

Is there an outlier here?

prestige = sm.datasets.get_rdataset("Duncan", "carData", cache=True).data
fig = plt.figure(figsize = (10, 7))
ax = plt.axes(projection ="3d")
ax.scatter3D(prestige.income, prestige.education, prestige.prestige, alpha=0.7)
plt.show()

Fitting the Duncan prestige data

prestige_model = smf.ols("prestige ~ income + education", data=prestige).fit()
print(prestige_model.summary())

                            OLS Regression Results                            
==============================================================================
Dep. Variable:               prestige   R-squared:                       0.828
Model:                            OLS   Adj. R-squared:                  0.820
Method:                 Least Squares   F-statistic:                     101.2
Date:                Tue, 01 Nov 2022   Prob (F-statistic):           8.65e-17
Time:                        14:02:05   Log-Likelihood:                -178.98
No. Observations:                  45   AIC:                             364.0
Df Residuals:                      42   BIC:                             369.4
Df Model:                           2                                         
Covariance Type:            nonrobust                                         
==============================================================================
                 coef    std err          t      P>|t|      [0.025      0.975]
------------------------------------------------------------------------------
Intercept     -6.0647      4.272     -1.420      0.163     -14.686       2.556
income         0.5987      0.120      5.003      0.000       0.357       0.840
education      0.5458      0.098      5.555      0.000       0.348       0.744
==============================================================================
Omnibus:                        1.279   Durbin-Watson:                   1.458
Prob(Omnibus):                  0.528   Jarque-Bera (JB):                0.520
Skew:                           0.155   Prob(JB):                        0.771
Kurtosis:                       3.426   Cond. No.                         163.
==============================================================================

Notes:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.

Residual plot for Duncan prestige data

plt.scatter(prestige_model.resid, prestige_model.predict())
plt.show()

Robust regression

robust = smf.rlm(
  "prestige ~ income + education", 
  data = prestige, 
  M = sm.robust.norms.HuberT()).fit()
print(robust.summary())

                    Robust linear Model Regression Results                    
==============================================================================
Dep. Variable:               prestige   No. Observations:                   45
Model:                            RLM   Df Residuals:                       42
Method:                          IRLS   Df Model:                            2
Norm:                          HuberT                                         
Scale Est.:                       mad                                         
Cov Type:                          H1                                         
Date:                Tue, 01 Nov 2022                                         
Time:                        14:02:06                                         
No. Iterations:                    18                                         
==============================================================================
                 coef    std err          z      P>|z|      [0.025      0.975]
------------------------------------------------------------------------------
Intercept     -7.1107      3.879     -1.833      0.067     -14.713       0.492
income         0.7015      0.109      6.456      0.000       0.489       0.914
education      0.4854      0.089      5.441      0.000       0.311       0.660
==============================================================================

If the model instance has been used for another fit with different fit parameters, then the fit options might not be the correct ones anymore .